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Gain of filter circuit?

  • #1
Thread moved from Electrical Engineering forum, so no template included.
E9LpVlj.png

The question wants to know the gain of the this circuit, as well as a bode plot for the frequency response, and what type of filter this is when C1 is much greater than C2. Could someone double check my answers and assist in the bode plot?

My attempt:

For the Gain:
0urkM66.png

If C1 is much greater than C2 I said the filter would be a low pass filter.

I'm not sure how to proceed with the Bode Plot.
 

Answers and Replies

  • #2
Henryk
Gold Member
251
97
Hi, I'm afraid your answer is not correct.
The simplest way to analyze the circuit is to note that ## C_1## and ##C_2 ## form a capacitive voltage divider connected to ## V_{IN} ##. The divider is equivalent of ## C_1## and ##C_2 ## in parallel connected to the voltage source ## V_{eq} = V_{IN} \frac {C_1} {C_1 + C_2} ##. Then connect the resistor between the output of the two capacitors and the ground. What you have is a high pass filter with the corner frequency given by ## \omega _c = R *(C_1+C_2)##
If ##C_1 \gg C_2 ##, then you can simply ignore the second capacitor. The equivalent circuits is showing you why you can do that.
I'm attaching a drawing of the equivalent circuit.
circuit.jpg
 
  • #3
Tom.G
Science Advisor
2,935
1,690
Won't there be a pole due to R1C2?
 
  • #4
NascentOxygen
Staff Emeritus
Science Advisor
9,243
1,068
  • #5
Tom.G
Science Advisor
2,935
1,690
Oops! You're right. No pole there. I just worked it out.
 

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